A function satisfies the equation f(x + y) = f(x) f(y) for all x, y . Suppose that the function is differentiable at x = 0 and f’(0) = 2. Prove that f’(x) = 2f(x).

Given f(x) is differentiable at x = 0 and f(x) ≠ 0

And f(x + y) = f(x)f(y) also f’(0) = 2

To prove: f’(x) = 2f(x)

As we know that,

f’(x) =

as f(x+h) = f(x)f(h)

∴ f’(x) =

⇒ f’(x) = …(1)

As

f(x + y) = f(x)f(y)

put x = y = 0

∴ f(0+0) = f(0)f(0)

⇒ f(0) = {f(0)}²

∴ f(0) = 1 {∵ f(x) ≠ 0 ….given}

∴ equation 1 is deduced as-

f’(x) =

⇒ f’(x) =

⇒ f’(x) = f(x)f’(0) {using formula of derivative}

∴ f’(x) = 2 f(x) …proved {∵ it is given that f’(0) = 2}